SaigaNoobie 66 Posted February 18, 2009 Report Share Posted February 18, 2009 Ok, I need some people with a little more practical math knowledge than myself. Specifically in the Volume of Gas department. .729" Diameter pipe with gas flowing at approximately 1200FPS will allow 3.4783 Cubic Feet Per second to pass through the hole. At 1200FPS, dwell time in a 19" Barrel is .001319 seconds for the projectile. I'm trying to calculate the pressure exerted by the gas, and thereby calculate how much gas really enters the Gas tube through three .072" diameter holes. One thing we will assume: the 19" barrel is .729" it's entire length, giving it a volume of 7.93 Cubic Inches, or .004589 Cubic Feet. So, with a time of 1.319e-3 seconds, and three .072" diameter holes, and the Pressure (P) that someone can show me how to calculate, I'd like to determine the precise amount of air pushed into the gas chamber. I may be missing some information, you guys might be able to help me out. Thanks~ Quote Link to post Share on other sites
SaigaNoobie 66 Posted February 18, 2009 Author Report Share Posted February 18, 2009 Hrm on second thought, maybe I need to calculate what Force is required to accelerate a 1 1/8oz projectile from 0 to 1200fps in 19", then using that Force, we can determine pressure? But then how do we go from pressure to volume per time through a certain diameter pipe? Quote Link to post Share on other sites
hallboss 1 Posted February 18, 2009 Report Share Posted February 18, 2009 PV=nRT P=pressure in atmospheres (1 atm =760mm/hg) V=volume in liters n=is the moles of your gas (molecules per gram) R=is the gas constant which is .821 T=is the temp your gas is in kelvin's Substitute your knowns and use algebra to figure our you unknown. If you have three before and after knowns and need to know what effect a change will have, then use the formula P1(V1)/n1(T1)=P2(V2)/n2(T2) I would do it for you, but I am heading into a meeting right now. If what I wrote appears to be written in Chinese, sorry, I'll try to help you out when I get back. Otherwise I hope this helps. Quote Link to post Share on other sites
SaigaNoobie 66 Posted February 18, 2009 Author Report Share Posted February 18, 2009 Chinese? No. Physical Chemistry? Yes. I passed PChem... but not with an A. I do remember this equation (Remembered by saying Pervnert - PV equals nrt) I'll give it a whirl and you can check my math. Quote Link to post Share on other sites
SaigaNoobie 66 Posted February 18, 2009 Author Report Share Posted February 18, 2009 Ok, I believe I know the initial numbers P1 = 1 atm V1 = .129957 L (Capacity of barrel) n1 = ? R= .821 T1= 297.15 (75* F or 24* C) Using those and PV=nrt we can calculate n as 5.327e-4 So n1 = 5.327e-4 Now let's combine equations? P1(V1)/n1(T1)=P2(V2)/n2(T2) 1 atm (.129957L)/5.327e-4M(297.15K) = P2(.129957L)/n2(T2) or 6.3175atm/MK = P2/n2(T2) So now, How do we determine the Temperature of the Burning Smokeless Powder? I searched for it to no avail. How do we determine the quantity and makeup of gasses released by 3 Drams of Smokeless powder in a Shotgun shell? Those are the missing components. Quote Link to post Share on other sites
hallboss 1 Posted February 18, 2009 Report Share Posted February 18, 2009 Ok, I believe I know the initial numbers P1 = 1 atm V1 = .129957 L (Capacity of barrel) n1 = ? R= .821 T1= 297.15 (75* F or 24* C) Using those and PV=nrt we can calculate n as 5.327e-4 So n1 = 5.327e-4 Now let's combine equations? P1(V1)/n1(T1)=P2(V2)/n2(T2) 1 atm (.129957L)/5.327e-4M(297.15K) = P2(.129957L)/n2(T2) or 6.3175atm/MK = P2/n2(T2) So now, How do we determine the Temperature of the Burning Smokeless Powder? I searched for it to no avail. How do we determine the quantity and makeup of gasses released by 3 Drams of Smokeless powder in a Shotgun shell? Those are the missing components. I had a break for a minute. The numbers look good. When I get a chance, I will look around for some pressure/temp gradients for burnt/burning powder. Quote Link to post Share on other sites
jrance@iacwds.com 716 Posted February 18, 2009 Report Share Posted February 18, 2009 DAYUM! We gots us some smart folks here! Quote Link to post Share on other sites
SaigaNoobie 66 Posted February 18, 2009 Author Report Share Posted February 18, 2009 Chemistry was my Major in college, Minor in Bio. But pushing myself at 18 credit hrs and 20 credit hrs my last two semesters really wore me thin. Full time is considered 12 credit hours, and more than 17 per semester isn't allowed.... unless you get the head of your Major department to sign off.... almost wish I had taken that extra half a semester to finish... damn that burnt me out. I could actually use the things I tried to learn in Pchem had I had the time to learn them all better Quote Link to post Share on other sites
hallboss 1 Posted February 18, 2009 Report Share Posted February 18, 2009 http://admin.pubs.acs.org/doi/pdf/10.1021/ie50257a006 That article was written in 1931 and toward the bottom it has a table that states smokeless powder burns from 2900*C-3500*C (dependant on chemical makeup). I would use 3000*C or 3273 K. Don't forget to do the enthalpy change of the air surrounding the reaction (I think it would be miniscule however). Now that I think of it, what would be the specific heat of air....0 probably, so nevermind. Quote Link to post Share on other sites
railman1 0 Posted February 19, 2009 Report Share Posted February 19, 2009 I read something like this thread and I am amazed at the depth of my ignorance. Quote Link to post Share on other sites
SaigaNoobie 66 Posted February 19, 2009 Author Report Share Posted February 19, 2009 I don't think the numbers are working. Can we calculate how much gas would be able to flow through a .144" diameter hole in .001319 seconds with 10,500 PSI (714 atm) of pressure behind it? Quote Link to post Share on other sites
Bounce12 407 Posted February 19, 2009 Report Share Posted February 19, 2009 (edited) I don't think the numbers are working. Can we calculate how much gas would be able to flow through a .144" diameter hole in .001319 seconds with 10,500 PSI (714 atm) of pressure behind it? I have a degree in mathematics, but it was 27 years ago... My brain has been in so many other places since then... However, if I can just clear my mind for a minute and look at this... All right, look, the initial numbers are what they are, but nowhere in that equation is the variable you are looking for. So solving for these variables won't help. I don't think the equation is complex enough. You actually have four holes, right (the barrel and three ports). The pressure will be variable depending on whether the slug is behind or in front of the ports. The gunpowder continues to burn while the slug travels down the barrel, so the temperature and pressure will continuously fluctuate. Neither is a static number, yet you have them as such in the equation. This equation won't work. Let me think about this. I'm watching Larry Kudlow and I need to finish the show, first. LOL I don't think that's the equation you want. Edited February 19, 2009 by Bounce12 Quote Link to post Share on other sites
Bounce12 407 Posted February 19, 2009 Report Share Posted February 19, 2009 We need to spend some time analyzing the stuff that goes in to this. You say the barrel is a certain length, but as the projectile moves down the barrel, the volume is changing continuously. Volume one would be the volume as the projectile passes the ports and volume 2 would be as it leaves the barrel, right? (just that short distance) Therefore, the dwell time is ONLY the time between the projectile passing the ports and leaving the barrel. The time it takes to get TO the ports is pretty meaningless, right? There will be SOME amount of cold air being blown in, but it will be lost in the static when the hot gasses enter AFTER the projectile passes the ports. I'll get back to this in an hour or so. Quote Link to post Share on other sites
Bounce12 407 Posted February 19, 2009 Report Share Posted February 19, 2009 (edited) The other thing is, are you counting high brass magnum shells or low brass skeet stuff. How do you presume to account for the difference? As I sit here multi-tasking, it occurred to me that algebra is not going to hack it. The volume changes over time, the pressure changes over time and the temperature could as well. Those will take integrals (calculus III) to solve. The time being the time from port passage to barrel evacuation. I think finding the EXACT answer is quite a bit more complicated. If close is good enough, pick a pressure and a temp value somewhere in the middle and wag it. Edited February 19, 2009 by Bounce12 Quote Link to post Share on other sites
gtnichols 51 Posted February 19, 2009 Report Share Posted February 19, 2009 Don't forget the thickness of the barrel and the angles of the holes. Quote Link to post Share on other sites
Bounce12 407 Posted February 19, 2009 Report Share Posted February 19, 2009 Don't forget the thickness of the barrel and the angles of the holes. Yeah, and there is going to be friction from the turbulence swirling in the chamber and back pressure from the spring and other friction in the mechanism as the piston moves back. If you want a no-shit EXACT answer, it will take quite a bit more effort just to set the variables up. It's not just a matter of how many molecules pass through three ports in T amount of time at P pressure. That would be a wild ass guess (wag) answer, but it will be far from exact. Quote Link to post Share on other sites
SaigaNoobie 66 Posted February 19, 2009 Author Report Share Posted February 19, 2009 i want a "within +- 10% answer" Quote Link to post Share on other sites
Bounce12 407 Posted February 19, 2009 Report Share Posted February 19, 2009 i want a "within +- 10% answer" All right, let me work on this. I'm going to have to remember how to do fucking integrals again. Man, that shit leaves your brain if you don't use it. When do you need the answer by? (please don't say tonight) Quote Link to post Share on other sites
Bounce12 407 Posted February 19, 2009 Report Share Posted February 19, 2009 How far is it from the ports to the end of the barrel? Are you shooting high brass or low brass and what temps and pressures do you want to assume? I think the pressure continues to build as the projectile is in the barrel. I don't even think all the powder burns before the projectile leaves the barrel. So that shit will be increasing the whole way. The speed of the projectile will be increasing the whole way and so using a constant projectile speed for the distance between the ports and the end of the barrel is not right. This is a very complicated question. Quote Link to post Share on other sites
Bounce12 407 Posted February 19, 2009 Report Share Posted February 19, 2009 To figure this we're going to need some more data. How much powder are you burning (low brass/high brass)? What rate of burn do you want to assume? This will change depending on the powder makeup, shape and size. How much remains unburned after the projectile leaves the barrel? Do you want to assume the powder burns at the same rate for the entire time (an assumption I wouldn't make, but I don't know what else to assume). What temp do you want to assume the powder burns at? The complexity of this question continues to grow. You're going to have to take a stab at some "assumptions." Quote Link to post Share on other sites
Bounce12 407 Posted February 19, 2009 Report Share Posted February 19, 2009 I just measured my S-12 from where I THINK the ports are to the end of the barrel and I get 10.5 inches. Is that close enough for a "distance?" Quote Link to post Share on other sites
Bounce12 407 Posted February 19, 2009 Report Share Posted February 19, 2009 Look Saiganoob, the more I think about this, the more complicated it gets. You're asking a tough question. The easiest way is to eliminate most of the variables as inconsequential and do some simple algebra. I'm not sure how close the answer will be, but you can come up with something in the right solar system. See if you can determine about what pressure is in a shotgun barrel and see how long it takes the projectile to go 10.5 inches. Use the muzzle velocity - close enough. That's the time the pressure is going through the ports. Muzzle velocity assumption is 1325 ft/sec That equals about 16000 inches per second. That equals about 0.00066 seconds from the ports to the end of the barrel. SO, T = 0.00066. That's a wild ass guess with many erroneous assumtions built in, but it's close enough. Now, what pressure do you want to assume is behind the projectile during that time? Quote Link to post Share on other sites
Bounce12 407 Posted February 19, 2009 Report Share Posted February 19, 2009 This guy says the average pressure in a 12-gauge is 11,500 psi. http://www.randywakeman.com/shotgun_pressures.htm I have no idea how correct that is. So, how much volume will go through three ports of the size you describe at 11,500 psi for 0.00066 seconds. You can do that! I can't do ALL your homework for you. that answer will be close enough Quote Link to post Share on other sites
Vultite 57 Posted February 19, 2009 Report Share Posted February 19, 2009 I'll try to pass this to a few people in my group of friends with various degrees that should be able to answer this, i'll let ya know when they do... Quote Link to post Share on other sites
hallboss 1 Posted February 19, 2009 Report Share Posted February 19, 2009 When you figure on the projectile moving, you create the variables of the pressure increasing as well as the volume increasing, combine that with the speed of the projectile, the equation is so complicated, you might as well factor in coeffecient of drag, gravity and friction to surface area. Not implicating I have the answers, but maybe another way to look at this is to solve for what pressure is needed to propel the projectile at your desired speed. Then after you have that, assume the projectile is fixed at the end of the barrel, now your volume is fixed. Plug in that pressure into the equation and solve for........ Hey Noobie, what are we solving for? Quote Link to post Share on other sites
Vultite 57 Posted February 19, 2009 Report Share Posted February 19, 2009 (edited) oh hey, I need more info man, where are the holes located on the length of the barrel ? Edited February 19, 2009 by Vultite Quote Link to post Share on other sites
Bounce12 407 Posted February 19, 2009 Report Share Posted February 19, 2009 oh hey, I need more info man, where are the holes located on the length of the barrel ? Mine are 10.5 inches from the muzzle. Quote Link to post Share on other sites
SaigaNoobie 66 Posted February 19, 2009 Author Report Share Posted February 19, 2009 (edited) LOL wow you all are knocking this shit out! Assume 10.5" from Gas Port holes to end of muzzle. Assume 1200 FPS Assume 10500 PSI Assume Constant speed (Acceleration is almost instantaneous when you're talking 6/10000ths of a second. In the time it takes for the projectile to go 10.5" at 1200 FPS, (7.23167e-4 seconds) How much gas will flow through a .144" diameter hole? I can do the math, i just need to find the equation that uses Pressure, Time, Diameter of hole to solve for Volume. Edited February 19, 2009 by SaigaNoobie Quote Link to post Share on other sites
Twinsen 86 Posted February 19, 2009 Report Share Posted February 19, 2009 You can just fill the barrel with water to get a volume reading... Quote Link to post Share on other sites
SaigaNoobie 66 Posted February 19, 2009 Author Report Share Posted February 19, 2009 LOL Twin. I can solve for volume with math. I know the volume of a barrel, what i don't know is how much Gas flows to the gas system in .0006 seconds under 10500 PSI through .144" Diameter hole. Quote Link to post Share on other sites
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